What is the integral of (3x^4)/((2x^5-5)^4) ?

The integral of (3x^4)/((2x^5-5)^4) is -1/(10(-5+2x^5)^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x^4)/((2x^5-5)^4)

\int \frac{3 x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

- \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}} + C

Solution steps

\int \frac{3 x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{10 u ^{4}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \frac{1}{10} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

= 3 \cdot \frac{1}{10} (- \frac{1}{3 u ^{3}})

Substitute back

u = - 5 + 2 x^{5}

= 3 \cdot \frac{1}{10} (- \frac{1}{3 ( - 5 + 2 x ^{5} ) ^{3}})

Simplify

3 \cdot \frac{1}{10} (- \frac{1}{3 ( - 5 + 2 x ^{5} ) ^{3}}) : - \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}}

= - \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}}

Add a constant to the solution

= - \frac{1}{10 ( - 5 + 2 x ^{5} ) ^{3}} + C

\int 6 t^{2} d t

d er i v a t i v e x^{3} e^{- 2 x}

\int (\frac{1}{x ^{2}} - x^{2} - \frac{1}{3}) d x

\int \frac{2 x}{x ^{2} - 2 x - 3} d x

\frac{d y}{d x} = e^{4 x}

What is the integral of (3x^4)/((2x^5-5)^4) ?
The integral of (3x^4)/((2x^5-5)^4) is -1/(10(-5+2x^5)^3)+C